Alternating Current, Voltage (rms and Average) Questions in English

(C) The instantaneous value of an alternating voltage is given by $V(t) = V_{\max} \sin(\omega t)$.
To find the average value over a complete cycle (from $t = 0$ to $t = T$),we integrate the function over the period $T$ and divide by the period:
$V_{\text{avg}} = \frac{1}{T} \int_{0}^{T} V_{\max} \sin(\omega t) dt$.
Since $\omega = \frac{2\pi}{T}$,the integral becomes:
$V_{\text{avg}} = \frac{V_{\max}}{T} \int_{0}^{T} \sin\left(\frac{2\pi}{T} t\right) dt$.
The integral of the sine function over a full period is zero because the positive area of the first half-cycle exactly cancels out the negative area of the second half-cycle.
Therefore,$V_{\text{avg}} = 0$.

(A) The instantaneous current is given by $i = i_{m} \sin(\omega t)$.
At the $rms$ value, $i = i_{rms} = \frac{i_{m}}{\sqrt{2}}$.
Substituting this into the equation: $\frac{i_{m}}{\sqrt{2}} = i_{m} \sin(\omega t)$.
$\sin(\omega t) = \frac{1}{\sqrt{2}}$, which implies $\omega t = \frac{\pi}{4}$.
Since $\omega = 2 \pi f$, we have $2 \pi f t = \frac{\pi}{4}$.
$t = \frac{1}{8f}$.
Given $f = 50 \text{ Hz}$, $t = \frac{1}{8 \times 50} = \frac{1}{400} \text{ s}$.
$t = 0.0025 \text{ s} = 2.5 \text{ ms}$.

(B) The instantaneous voltage of an $AC$ generator is given by $V = V_m \sin(\omega t)$, where $V_m$ is the peak voltage and $\omega = 2 \pi \nu$ is the angular frequency.
Given: $V_m = 100 \text{ V}$, $V = 2 \text{ V}$ at $t = \frac{1}{100 \pi} \text{ s}$.
Substituting these values into the equation:
$2 = 100 \sin\left(2 \pi \nu \times \frac{1}{100 \pi}\right)$
$2 = 100 \sin\left(\frac{\nu}{50}\right)$
$\frac{2}{100} = \sin\left(\frac{\nu}{50}\right)$
Since the angle $\frac{\nu}{50}$ is very small, we can use the approximation $\sin(\theta) \approx \theta$:
$\frac{1}{50} = \frac{\nu}{50}$
$\nu = 1 \text{ Hz}$.

(B) Given: $I_{rms} = 5 \text{ A}$,frequency $f = 50 \text{ Hz}$.
At $t=0$,$I=0$,so the equation for current is $I = I_0 \sin(\omega t)$.
First,calculate the peak current $I_0 = \sqrt{2} \times I_{rms} = 5\sqrt{2} \text{ A}$.
Next,calculate the angular frequency $\omega = 2\pi f = 2\pi \times 50 = 100\pi \text{ rad/s}$.
Now,substitute $t = \frac{1}{300} \text{ s}$ into the equation:
$I = 5\sqrt{2} \sin(100\pi \times \frac{1}{300})$
$I = 5\sqrt{2} \sin(\frac{\pi}{3})$
Since $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$,we get:
$I = 5\sqrt{2} \times \frac{\sqrt{3}}{2} = 5 \sqrt{\frac{2 \times 3}{4}} = 5 \sqrt{\frac{3}{2}} \text{ A}$.

(C) The given equation is $V = V_0 \sin(\omega t)$,where $V_0 = 311 \ V$ and $\omega = 314 \ rad/s$.
The $rms$ value of voltage is given by $V_{rms} = \frac{V_0}{\sqrt{2}}$.
$V_{rms} = \frac{311}{1.414} \approx 220 \ V$.
The angular frequency is $\omega = 2\pi f$.
$314 = 2 \times 3.14 \times f$.
$314 = 6.28 \times f$.
$f = \frac{314}{6.28} = 50 \ Hz$.
Thus,the $rms$ voltage is $220 \ V$ and the frequency is $50 \ Hz$.

(B) The given equation is $i = i_1 \sin \omega t + i_2 \cos \omega t$.
We can rewrite this as $i = A \sin(\omega t + \phi)$,where $A$ is the amplitude of the resultant current.
To find $A$,we compare the coefficients: $i_1 = A \cos \phi$ and $i_2 = A \sin \phi$.
Squaring and adding these equations: $i_1^2 + i_2^2 = A^2 \cos^2 \phi + A^2 \sin^2 \phi = A^2(\cos^2 \phi + \sin^2 \phi) = A^2$.
Thus,the amplitude $A = \sqrt{i_1^2 + i_2^2}$.
The root mean square (rms) value of a sinusoidal current $i = A \sin(\omega t + \phi)$ is given by $i_{rms} = \frac{A}{\sqrt{2}}$.
Substituting the value of $A$,we get $i_{rms} = \sqrt{\frac{i_1^2 + i_2^2}{2}}$.

(D) The instantaneous current is given by $ I = I_0 \sin(\omega t) $.
At the $ rms $ value, $ I = I_{rms} = \frac{I_0}{\sqrt{2}} $.
So, $ \frac{I_0}{\sqrt{2}} = I_0 \sin(\omega t_1) \Rightarrow \sin(\omega t_1) = \frac{1}{\sqrt{2}} $.
This gives $ \omega t_1 = \frac{\pi}{4} $.
Since $ \omega = \frac{2\pi}{T} $, we have $ \frac{2\pi}{T} t_1 = \frac{\pi}{4} \Rightarrow t_1 = \frac{T}{8} $.
The peak value occurs at $ t_2 = \frac{T}{4} $.
The time taken to reach the peak value from the $ rms $ value is $ \Delta t = t_2 - t_1 = \frac{T}{4} - \frac{T}{8} = \frac{T}{8} $.
Given frequency $ f = 50 \,Hz $, the time period is $ T = \frac{1}{f} = \frac{1}{50} \,s = 0.02 \,s $.
Therefore, $ \Delta t = \frac{0.02}{8} = 0.0025 \,s = 2.5 \times 10^{-3} \,s $.
Thus, option $ D $ is correct.

(B) In India,including Karnataka,the standard domestic $AC$ power supply is specified as $220 \text{ V}, 50 \text{ Hz}$.
By convention,the voltage value provided for $AC$ circuits is the Root Mean Square $(RMS)$ value,as it represents the effective voltage that would produce the same heating effect as an equivalent $DC$ voltage.
The frequency of $50 \text{ Hz}$ represents the number of cycles per second of the alternating current.
Therefore,$220 \text{ V}$ is the $RMS$ voltage and $50 \text{ Hz}$ is the frequency.

(C) The $ AC $ current is represented by the equation $ I = I_{\text{max}} \sin(\omega t) $.
For a complete cycle,the average value of current $ I_{\text{avg}} $ is defined as the integral of current over the time period $ T $ divided by the time period $ T $.
$ I_{\text{avg}} = \frac{1}{T} \int_{0}^{T} I_{\text{max}} \sin(\omega t) dt $.
Since the integral of $ \sin(\omega t) $ over a complete period $ T $ is zero,the average current over a complete cycle is always $ 0 $.
Therefore,for any sinusoidal $ AC $ current,the average value over a complete cycle is $ 0 $.

(C) The root mean square $(RMS)$ value of voltage,$ V_{rms} $,and the peak voltage,$ V_{0} $,of an $A$.$C$. source are related by the formula:
$ V_{rms} = \frac{V_{0}}{\sqrt{2}} $
Given that the multimeter reads the $RMS$ value,$ V_{rms} = 100 \,V $.
To find the peak voltage $ V_{0} $,we rearrange the formula:
$ V_{0} = V_{rms} \times \sqrt{2} $
Substituting the given value:
$ V_{0} = 100 \times 1.414 = 141.4 \,V $
Thus,the peak value of the voltage of the $A$.$C$. source is $ 141.4 \,V $.

(D) Given: $R = 20 \Omega$,$V = 200 \sin (10 \pi t)$.
Using Ohm's law,$I = \frac{V}{R} = \frac{200}{20} \sin (10 \pi t) = 10 \sin (10 \pi t)$.
The peak current is $I_0 = 10 \text{ A}$.
The rms current is $I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \text{ A}$.
At peak value,$10 = 10 \sin (10 \pi t_1) \Rightarrow \sin (10 \pi t_1) = 1 \Rightarrow 10 \pi t_1 = \frac{\pi}{2} \Rightarrow t_1 = \frac{1}{20} \text{ s}$.
At rms value,$\frac{10}{\sqrt{2}} = 10 \sin (10 \pi t_2) \Rightarrow \sin (10 \pi t_2) = \frac{1}{\sqrt{2}} \Rightarrow 10 \pi t_2 = \frac{\pi}{4} \Rightarrow t_2 = \frac{1}{40} \text{ s}$.
The time taken to change from peak to rms is $\Delta t = t_1 - t_2 = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} = 0.025 \text{ s} = 2.5 \times 10^{-2} \text{ s}$.

(C) The given alternating current is $i = 3 \sin \omega t + 4 \cos \omega t$.
We can express this in the form $i = I_0 \sin(\omega t + \phi)$,where $I_0$ is the peak current.
Using the identity $a \sin \theta + b \cos \theta = \sqrt{a^2 + b^2} \sin(\theta + \phi)$,we get:
$I_0 = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \ A$.
The $rms$ current is given by the formula $I_{rms} = \frac{I_0}{\sqrt{2}}$.
Substituting the value of $I_0$,we get $I_{rms} = \frac{5}{\sqrt{2}} \ A$.

(A) The given voltage $V = 200 V$ is the $RMS$ voltage $(V_{rms})$.
Peak voltage is given by the formula $V_{peak} = \sqrt{2} \times V_{rms}$.
Substituting the values,$V_{peak} = 1.414 \times 200 = 282.8 V$.
Peak current is given by $I_{peak} = \frac{V_{peak}}{R}$.
Substituting the values,$I_{peak} = \frac{282.8}{280} \approx 1.01 A$.
Therefore,the peak current is nearly $1 A$.

(B) The $rms$ value of $AC$ is given as $V_{rms} = 220 \ V$.
The peak value $(V_0)$ of $AC$ is calculated as $V_0 = \sqrt{2} \times V_{rms} = 1.414 \times 220 \approx 311 \ V$.
For $DC$,the voltage remains constant at $220 \ V$.
Since the peak voltage of $220 \ V$ $AC$ is $311 \ V$,which is significantly higher than the constant $220 \ V$ of $DC$,$220 \ V$ $AC$ is more dangerous than $220 \ V$ $DC$.

(B) The given emf is $E = 6 \sin \omega t + 4 \sin 2 \omega t \text{ V}$.
For a non-sinusoidal periodic waveform $E = E_1 \sin \omega t + E_2 \sin 2 \omega t + \dots$,the rms value is given by $E_{\text{rms}} = \sqrt{\frac{E_1^2 + E_2^2 + \dots}{2}}$.
Here,$E_1 = 6 \text{ V}$ and $E_2 = 4 \text{ V}$.
Substituting these values into the formula:
$E_{\text{rms}} = \sqrt{\frac{6^2 + 4^2}{2}}$
$E_{\text{rms}} = \sqrt{\frac{36 + 16}{2}}$
$E_{\text{rms}} = \sqrt{\frac{52}{2}}$
$E_{\text{rms}} = \sqrt{26} \text{ V}$.
Thus,the rms value of the emf is $\sqrt{26} \text{ V}$.

(A) The given emf is $E = 8 \sin \omega t + 6 \cos \omega t$.
We can express this in the form $E = A \sin(\omega t + \phi)$,where $A$ is the peak amplitude.
Comparing $8 \sin \omega t + 6 \cos \omega t$ with $A \sin(\omega t + \phi) = A \sin \omega t \cos \phi + A \cos \omega t \sin \phi$,we get:
$A \cos \phi = 8$ and $A \sin \phi = 6$.
Squaring and adding these equations:
$A^2(\cos^2 \phi + \sin^2 \phi) = 8^2 + 6^2$
$A^2 = 64 + 36 = 100$
$A = 10 \text{ V}$.
The rms value $E_{rms}$ is related to the peak value $A$ by the formula $E_{rms} = \frac{A}{\sqrt{2}}$.
$E_{rms} = \frac{10}{\sqrt{2}} = \frac{10 \times \sqrt{2}}{2} = 5 \sqrt{2} \text{ V}$.

(A) The given emf is $E = 8 \sin \omega t + 6 \cos \omega t$.
We can rewrite this in the form $E = E_0 \sin(\omega t + \phi)$.
To do this,multiply and divide by $\sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.
$E = 10 \left( \frac{8}{10} \sin \omega t + \frac{6}{10} \cos \omega t \right)$.
Let $\cos \phi = \frac{8}{10} = 0.8$ and $\sin \phi = \frac{6}{10} = 0.6$.
Then $E = 10 (\sin \omega t \cos \phi + \cos \omega t \sin \phi) = 10 \sin(\omega t + \phi)$.
Comparing this with $E = E_0 \sin(\omega t + \phi)$,we get the peak value $E_0 = 10 \text{ V}$.
The $RMS$ value is given by $E_{RMS} = \frac{E_0}{\sqrt{2}}$.
$E_{RMS} = \frac{10}{\sqrt{2}} = 5 \sqrt{2} \text{ V}$.

(B) The instantaneous voltage $V$ is given by $V = V_0 \sin(\omega t)$,where $V_0$ is the peak voltage and $\omega = 2\pi f$.
Given $f = 50 \ Hz$,so $\omega = 2 \times \pi \times 50 = 100\pi \ rad/s$.
We want to find the time $t$ when $V = \frac{V_0}{2}$.
Substituting this into the equation: $\frac{V_0}{2} = V_0 \sin(100\pi t)$.
$\sin(100\pi t) = \frac{1}{2}$.
Since $\sin(30^\circ) = \frac{1}{2}$,we have $100\pi t = \frac{\pi}{6}$.
$t = \frac{\pi}{6 \times 100\pi} = \frac{1}{600} \ s$.

(B) The standard equation for an alternating current is given by $I(t) = I_0 \sin(\omega t)$,where $I_0$ is the peak current and $\omega$ is the angular frequency.
Comparing this with the given equation $I(t) = 50 \sin(200 \pi t)$,we find the peak current $I_0 = 50 \text{ A}$ and the angular frequency $\omega = 200 \pi \text{ rad/s}$.
The $RMS$ value of the current is calculated as $I_{RMS} = \frac{I_0}{\sqrt{2}} = \frac{50}{\sqrt{2}} = 25 \sqrt{2} \text{ A}$.
The frequency $f$ is related to the angular frequency by the formula $\omega = 2 \pi f$.
Substituting the value of $\omega$,we get $200 \pi = 2 \pi f$,which simplifies to $f = 100 \text{ Hz}$.
Thus,the frequency is $100 \text{ Hz}$ and the $RMS$ value is $25 \sqrt{2} \text{ A}$.

(D) The instantaneous current in an $A.C.$ circuit is given by $i = i_0 \sin(\omega t + \phi)$.
At maximum value,$i = i_0$,which occurs at $\omega t_1 = \frac{\pi}{2}$.
At $R.M.S.$ value,$i = i_{R.M.S.} = \frac{i_0}{\sqrt{2}}$,which occurs at $\omega t_2 = \frac{\pi}{4}$ (or $\frac{3\pi}{4}$).
The time interval $\Delta t = t_1 - t_2$ is given by $\omega \Delta t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.
Since $\omega = 2 \pi f$,we have $2 \pi f \Delta t = \frac{\pi}{4}$.
Substituting $f = 50 Hz$,we get $2 \pi (50) \Delta t = \frac{\pi}{4}$.
$100 \pi \Delta t = \frac{\pi}{4} \Rightarrow \Delta t = \frac{1}{400} s$.
$\Delta t = 0.0025 s = 2.5 ms$.

(A) The given equation is $i = 2 \sin \omega t + 6 \cos \omega t$.
This is of the form $i = a \sin \omega t + b \cos \omega t$,where $a = 2$ and $b = 6$.
The peak current $i_0$ is given by $i_0 = \sqrt{a^2 + b^2}$.
$i_0 = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2 \sqrt{10} \ A$.
The $R.M.S.$ current $i_{rms}$ is related to the peak current $i_0$ by the formula $i_{rms} = \frac{i_0}{\sqrt{2}}$.
$i_{rms} = \frac{2 \sqrt{10}}{\sqrt{2}} = 2 \sqrt{5} \ A$.

(A) The root mean square voltage is given as $V_{rms} = 220 \ V$.
The peak voltage $V_0$ is related to $V_{rms}$ by the formula $V_0 = V_{rms} \sqrt{2}$.
Therefore, $V_0 = 220 \sqrt{2} \ V$.
The angular frequency $\omega$ is given by $\omega = 2 \pi f$, where $f = 50 \ \text{Hz}$.
Thus, $\omega = 2 \pi \times 50 = 100 \pi \ \text{rad/s}$.
The instantaneous potential difference is given by $V(t) = V_0 \cos(\omega t)$.
Substituting the values, we get $V(t) = 220 \sqrt{2} \cos(100 \pi t)$.

(D) The root mean square (rms) value of a periodic function $v(t)$ is defined as $V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2(t) dt}$.
From the given figure,the potential difference $v(t)$ is $V_0$ for the interval $0 \le t < \frac{T}{2}$ and $0$ for the interval $\frac{T}{2} \le t < T$.
Thus,$V_{rms}^2 = \frac{1}{T} \left[ \int_{0}^{T/2} V_0^2 dt + \int_{T/2}^{T} 0^2 dt \right]$.
$V_{rms}^2 = \frac{1}{T} \left[ V_0^2 \cdot \frac{T}{2} + 0 \right] = \frac{V_0^2}{2}$.
Taking the square root on both sides,we get $V_{rms} = \frac{V_0}{\sqrt{2}}$.

(C) The given equation for alternating current is $I = I_0 \sin(\omega t + \phi)$,where $I_0 = 20 \ A$ and $\omega = 100 \pi \ rad/s$.
The root mean square (r.m.s.) value of the current is given by $I_{rms} = \frac{I_0}{\sqrt{2}}$.
Substituting the value of $I_0$: $I_{rms} = \frac{20}{\sqrt{2}} = 10 \sqrt{2} \ A$.
The angular frequency is $\omega = 2 \pi f = 100 \pi$.
Solving for frequency $f$: $f = \frac{100 \pi}{2 \pi} = 50 \ Hz$.
Therefore,the r.m.s. value is $10 \sqrt{2} \ A$ and the frequency is $50 \ Hz$.

(C) Given: $I_{rms} = 10 \, A$ and $R = 12 \, \Omega$.
The peak current $I_0$ is given by the relation $I_0 = I_{rms} \times \sqrt{2}$.
$I_0 = 10 \times 1.414 = 14.14 \, A$.
The maximum potential difference $V_0$ across the resistor is given by Ohm's law: $V_0 = I_0 \times R$.
$V_0 = 14.14 \times 12 = 169.68 \, V$.

(D) The root mean square (r.m.s) current is defined as $i_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} i^2 dt}$.
Given $i = i_{0}(t / T)$,we have $i^2 = i_{0}^2 (t^2 / T^2)$.
Substituting this into the formula:
$i_{rms}^2 = \frac{1}{T} \int_{0}^{T} \frac{i_{0}^2 t^2}{T^2} dt = \frac{i_{0}^2}{T^3} \int_{0}^{T} t^2 dt$.
Evaluating the integral: $\int_{0}^{T} t^2 dt = \left[ \frac{t^3}{3} \right]_{0}^{T} = \frac{T^3}{3}$.
Therefore,$i_{rms}^2 = \frac{i_{0}^2}{T^3} \cdot \frac{T^3}{3} = \frac{i_{0}^2}{3}$.
Taking the square root,we get $i_{rms} = \frac{i_{0}}{\sqrt{3}}$.

(D) The relationship between maximum (peak) voltage $V_0$ and root mean square (rms) voltage $V_{rms}$ for a sinusoidal $AC$ circuit is given by $V_{rms} = \frac{V_0}{\sqrt{2}}$.
Rearranging this formula,we get the ratio of maximum voltage to $rms$ voltage as $\frac{V_0}{V_{rms}} = \sqrt{2}$.
Given that $\sqrt{2} \approx 1.414$,we convert this ratio into a percentage by multiplying by $100\%$.
Therefore,the percentage value is $1.414 \times 100\% = 141.4\%$.

(D) The power $P$ is given by the formula $P = V_{rms} \times I_{rms}$.
Given $P = 12 \text{ W}$ and $V_{rms} = 24 \text{ V}$,we can calculate the root mean square current $I_{rms}$ as:
$I_{rms} = \frac{P}{V_{rms}} = \frac{12}{24} = 0.5 \text{ A}$.
The peak current $I_0$ is related to the $I_{rms}$ by the formula $I_0 = I_{rms} \times \sqrt{2}$.
Substituting the value of $I_{rms}$,we get:
$I_0 = 0.5 \times \sqrt{2} = \frac{1}{2} \times \sqrt{2} = \frac{1}{\sqrt{2}} \text{ A}$.

(D) The induced emf in an $AC$ generator is given by the equation $\varepsilon = \varepsilon_0 \sin(\omega t)$.
At $t = 0$,$\varepsilon = \varepsilon_0 \sin(0) = 0$.
The emf reaches its maximum value when $\sin(\omega t) = 1$.
This occurs when $\omega t = \frac{\pi}{2}$.
Solving for $t$,we get $t = \frac{\pi}{2\omega}$.
Therefore,the induced emf is maximum at time $t = \frac{\pi}{2\omega}$.

(B) The instantaneous current is given by the equation $I = I_0 \sin(\omega t)$,where $I_0$ is the peak current and $\omega$ is the angular frequency.
To reach the peak value,the sine function must be equal to $1$,which occurs when the phase angle $\omega t = \frac{\pi}{2}$.
Substituting $\omega = 2 \pi f$,we get $2 \pi f t = \frac{\pi}{2}$.
Solving for time $t$,we get $t = \frac{1}{4f}$.
Given the frequency $f = 60 \text{ Hz}$,we substitute this value into the equation:
$t = \frac{1}{4 \times 60} = \frac{1}{240} \text{ s}$.
Therefore,the current takes $1/240 \text{ s}$ to reach its peak value from zero.